Static arrays at compile time in C++

Question

I was wondering if its possible to make the following answer more generic, in the sense that the type of the array be templated instead of just unsigned:

I've enclosed the whole thing in a struct like so:

template<typename ArrayType>
struct Array
{
template<ArrayType... args> struct ArrayHolder {
    static const ArrayType data[sizeof...(args)];
};

template<ArrayType... args> 
const ArrayType ArrayHolder<args...>::data[sizeof...(args)] = { args... };

template<size_t N, template<size_t> class F, ArrayType... args> 
struct generate_array_impl {
    typedef typename generate_array_impl<N-1, F, F<N>::value, args...>::result result;
};

template<template<size_t> class F, ArrayType... args> 
struct generate_array_impl<0, F, args...> {
    typedef ArrayHolder<F<0>::value, args...> result;
};

template<size_t N, template<size_t> class F> 
struct generate_array {
    typedef typename generate_array_impl<N-1, F>::result result;
};
};

but I get the following errors:

c++-4.6 -std=c++0x -o test test.cpp
test.cpp:49:17: error: specializing member ‘Array<ArrayType>::ArrayHolder<args>::data’ requires ‘template<>’ syntax

When showing compiler errors and code, it is usually quite helpfull to point in the code the exact line that the compiler is complaining about (i.e. which is line 49?) — David Rodríguez - dribeas, Aug 25 '11 at 11:20

score 8 · Accepted Answer · answered Aug 25 '11 at 10:32

It helps if you indent the struct. The problem is that you are defining the data static member variable inside the Array struct. But it should be at namespace scope:

template<typename ArrayType>
struct Array
{
    template<ArrayType... args> struct ArrayHolder {
        static const ArrayType data[sizeof...(args)];
    };

    template<size_t N, template<size_t> class F, ArrayType... args> 
        struct generate_array_impl {
            typedef typename generate_array_impl<N-1, F, F<N>::value, args...>::result result;
        };

    template<template<size_t> class F, ArrayType... args> 
        struct generate_array_impl<0, F, args...> {
            typedef ArrayHolder<F<0>::value, args...> result;
        };

    template<size_t N, template<size_t> class F> 
        struct generate_array {
            typedef typename generate_array_impl<N-1, F>::result result;
        };
};

template<typename ArrayType> template<ArrayType... args> 
        const ArrayType Array<ArrayType>::ArrayHolder<args...>::data[sizeof...(args)] = { args... };

score -2 · Answer 2 · answered Aug 25 '11 at 09:45

-2

Erm... Why don't you just use std::vector? Or if you want a non-resiable array, then use the equivalent type in boost?

answered Aug 25 '11 at 09:45

Arafangion

11,517
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score -2 · Answer 3 · answered Aug 25 '11 at 10:38

-2

I don't think the expression like 'template ' is a valid usage of C++11 variadic templates.

It can be of two forms as I understand:

template <typename... args> : variable number of generic type parameters
template <int... args> : variable number of integer(non-type) parameters

Refer to this wikipedia article for variadic templates.

answered Aug 25 '11 at 10:38

jj1

155
1
7

3

Those are template template arguments and they are valid. – Sebastian Mach Aug 25 '11 at 10:51

Static arrays at compile time in C++

3 Answers3

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